Let y = y(x) be the solution of the differential equation sec 2 x d x + ( e 2 y tan 2 x + tan x ) d y = 0 , 0 < x < 2 , y ( 4 ) = 0 . If y ( 6 ) = , then e 8 is equal to __________. [2024]

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Solution

Q.
Let y = y(x) be the solution of the differential equation $\sec^{2} x d x + (e^{2 y} \tan^{2} x + \tan x) d y = 0$ , $0 < x < \frac{π}{2}$ , $y (\frac{π}{4}) = 0$ . If $y (\frac{π}{6}) = α$ , then $e^{8 α}$ is equal to __________. [2024]

Ans.
(9)

We have, $\sec^{2} x d x + (e^{2 y} \tan^{2} x + \tan x) d y = 0$

Put $\tan x = t$

$\Rightarrow \sec^{2} x \frac{d x}{d y} = \frac{d t}{d y}$

Now, $\sec^{2} x \frac{d x}{d y} + (e^{2 y} \tan^{2} x + \tan x) = 0$

$\Rightarrow \frac{d t}{d y} + e^{2 y} t^{2} + t = 0 \Rightarrow \frac{d t}{d y} + t = - t^{2} e^{2 y}$

$\frac{1}{t^{2}} \frac{d t}{d y} + \frac{1}{t} = - e^{2 y}$

Again put $\frac{1}{t} = u \Rightarrow \frac{- 1}{t^{2}} \frac{d t}{d y} = \frac{d u}{d y}$

$∴ - \frac{d u}{d y} + u = - e^{2 y} \Rightarrow \frac{d u}{d y} - u = e^{2 y}$

$∴ I . F . = e^{- \int d y} = e^{- y}$

$∴ u e^{- y} = \int e^{- y} e^{2 y} d y$

$\Rightarrow u e^{- y} = \int e^{y} d y \Rightarrow \frac{1}{t} e^{- y} = e^{y} + c$

$\Rightarrow \frac{1}{\tan x} e^{- y} = e^{y} + c$ ... (i)

When, $x = \frac{π}{4}$ , y = 0

$∴ \frac{1}{\tan (\frac{π}{4})} e^{- 0} = e^{0} + c \Rightarrow 1 = 1 + c \Rightarrow c = 0$

Also, when $x = \frac{π}{6}, y = α$

$∴$ From (i), we have

$\frac{1}{\tan \frac{π}{6}} e^{- α} = e^{α} + 0 \Rightarrow \sqrt{3} = e^{2 α} \Rightarrow {(e^{2 α})}^{4} = {(\sqrt{3})}^{4}$

$\Rightarrow e^{8 α} = 9$ .

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